Show that the middle term in the expansion of $\left(x-\frac{1}{x}\right)^{2 n}$ is
$\frac{1 \times 3 \times 5 \times \ldots \times(2 n-1)}{n !} \times(-2)^{n}$
Given, expression is $\left(x-\frac{1}{x}\right)^{2 n}$
Since the index is $2 n$, which is even. So, there is only one middle term, i.e.,
$\left(\frac{2 n}{2}+1\right)$ th term $=(n+1)$ th term
$T_{n+1}={ }^{2 n} C_{n}(x)^{2 n-n}\left(-\frac{1}{x}\right)^{n}={ }^{2 n} C_{n}(-1)^{n}=(-1)^{n} \frac{(2 n !)}{n ! \cdot n !}$
$=(-1)^{n} \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \ldots(2 n-1) \cdot(2 n)}{n ! \cdot n !}=(-1)^{n} \frac{[1 \cdot 3 \cdot 5 \ldots(2 n-1)] \cdot[2 \cdot 4 \cdot 6 \ldots(2 n)]}{(1 \cdot 2 \cdot 3 \ldots n) \cdot n !}$
$=(-1)^{n} \frac{[1 \cdot 3 \cdot 5 \ldots(2 n-1)] \cdot 2^{n}[1 \cdot 2 \cdot 3 \ldots n]}{(1 \cdot 2 \cdot 3 \ldots n) \cdot n !}=(-2)^{n} \frac{[1 \cdot 3 \cdot 5 \ldots(2 n-1)] \cdot 2^{n}}{n !}$
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